424 RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13 Hint: One can write =-if I3 (X2 V2 / 2r + X3 V2rldAfi3 = I3 2 02 2/32 Also show that 13 = 2/2 13-12. Specialize Equations (13-84) and (13-85) for the case where the cross section is symmetrical with respect to the X2 axis. Utilize JjHe(X2, x3)HNo( 2, x3)dA = 0 where He is an even function and Ho an odd function of x3. Evaluate the co-efficients for the channel section of Example 13-5. Finally, specialize the equations for a doubly symmetric section. 13-13. Specialize (13-88) for a doubly symmetrical cross section. Then specialize further for negligible transverse shear deformation due to flexure and warping. The symmetry reductions are .X2= = 0 X-C2r= X3r == 03 P2 =3 =/3=t 0 1/A2 3 = 0 72 = 173 = 'i = 17i = 0 13-14. Consider the two following problems involving doubly symmetric cross section. (a) Establish "linearized" incremental equations by operating on (13-88)and retaining only linear ternms in the displacement increments. Specialize for a doubly symmetric cross section (see Prob. 13-12). (b) Consider the case where the cross section is doubly symmetric and the initial state is pure compression (Ft = -'). Determine the critical load with respect to torsional buckling for the following boundary conditions: 1. o =f = 0 at x = 0, L (restrained warping) df 02. o=1--=0 at = , L (unrestrained warping)dX Neutral equilibrium (buckling) is defined as the existence of a nontrivial solution of the linearized incremental equations for the same external load. One sets F1 = -P t2 = U3 = 01 = (02 = (03 = .f = and determines the value of P for which a nontrivial solution which satisfies the boundary conditions is possible. Employ the notation introduced in Example 13-7. 13-15. Determine the form of V, the strain energy density function (strain energy per unit length along the centroidal axis), expressed in terms of displace-ments. Assume no initial strain but allow for geometric nonlinearity. Note that V = V* when there is no initial strain. 14 Planar Deformation of a i1Planar Member 1I4 14-1. INTRODUCTION: GEOMETRICAL RELATIONS A member is said to be planar if--1. The centroidal axis is a plane curve. 2. The plane containing the centroidal axis also contains one of the principal inertia axes for the cross section. 3. The shear center axis coincides with or is parallel to the centroidal axis. However, the present discussion will be limited to the case where the shear center axis lies in the plane containing the centroidal axis. We consider the centroidal axis to be'defined with respect to a global reference frame having directions X1 and X2. This is shown in Fig. 14-1. The orthogonal unit vectors defining the orientation of the local fame (Y,, Y2) at a point are denoted by Ft, 2, where t[ points in the positive tangent direction and t x t2 = t3. Item 2 requires Y2 to be a principal inertia axis for the cross section. X2 ij Yl A I 12 1 .II4 -lFig. 14-1. Geometrical notation for plane curve. 425---- 426 427 PLANAR DEFORMATION OF A PLANAR MEMBER CHAP. 14 By definition, t dr dxl_ dx2 dS dSS 12 (14-1)t =--dS = dS +ldS Since we are taking t2 according to t1 x t2= 3, it follows that dx2 dxl, 2 dS 11 + 12 (14-2)dS dS The differentiation formulas for the unit vectors are dS dt2 = R t2 1 (14-3) dS = tlR where 1 dtl -dd2 . t x2 d2 x2 dxl R = d-S-dS + dS dS According to this definition, R is negative when dil/dS points in the negativet2 direction, e.g., for segment AB in Fig. 14-1. One could take 2 = ii, the unit normal vector defined by I dtl dt dS (14-4) dCSI rather than according to t x t2 = 3 but this choice is inconvenient whenthere is a reversal in curvature. Also, this definition degenerates at an inflection point, i.e., when dF/dS = . If the sense of the curvature is constant, one canalways orient the X1-X2 frame so that 2 coincides with hi,to avoid workingwith a negative R. To complete the geometrical treatment, we consider the general parametricrepresentation for the curve defining the centroidal axis, X = l(Y) t1-J1 x2 = x2(Y) where y is a parameter. The differential arc length is related to dy by dS = + dX ) + ( 2 12 dv = dy (14-6) According to this definition, the +S sense coincides with the direction of t We summarize here for convenience the essential geometric relations for a plane curve which are developed in Chapter 4. SEC. 14-2. FORCE-EQUILIBRIUM EQUATInm_-n increasing y. Using (14-6), the expressions for ti, 2, and 1/R in terms of v are t =1 dxi a-kTT + d-2 )V I d-x2-a (14-7) R o -/Jo-v 1 d2x d2 d22 dxl a03 --d d-d-2 d; A planar member subjected to in-plane forces (Xi-X2 plane for our notation)will experience only in-plane deformation. In what follows, we develop the governing equations for planar deformation of a arbitrary planar member.This formulation is restricted to the linear geometric case. The two basic solution procedures, namely, the displacement and force methods, are describedand applied to a circular member. We also present a simplified formulation (Marguerre's equations) which isvalid for a shallow member. Finally, we include a discussion of numericalintegration techniques, since one must resort to numerical integration whenthe cross section is not constant. 14-2. FORCE-EQUILIBRIUM EQUATIONS The notation associated with a positive normal cross section, i.e., a crosssection whose outward normal points in the + S direction, is shownin Fig. 14-2. We use the same notation as for the prismatic case, except that now the vector ,, 4 Y3 -Y Fig. 14-2. Force and moment components acting on a positive cross section.-- 428 PLANAR DEFORMATION OF A PLANAR MEMBER CHAP. 14 SEC. 14-3. PRINCIPLE OF VIRTUAL FORCES 429 components are with respect to the local frame (Y1, Y2, basic frame (X1 , X2, X3). The cross-sectional properties Y3) rathe are define r than the ed by We expand b and i in terms of the unit vectors for the local frame: Since Y2, that A = ff dY' =dy dy = dA 13 = ff(y2)2 dA 12 = |f'(y3)2 1A Y3 pass through the centroid and are principal directions, fJy 2 cA = d'y3dA = JY2Y3 dA = 0 (14-8) it follows (14-9) b = bl/l + b 212 tt-= mt 3 (14-13) Introducing the component expansionsin (14-12), and using the differentiation formulas for the unit vectors (.14-3), lead to the following scalar differential equilibrium equations: IF I F2F_1_ t-+ b = 0 When the member is planar (Xl-X2 plane) and is -subjected to a planar dS R loading, F3 = M1 = M2 = 0 (14-10) dF 2 dS F1 R (14-14) In this case, we work with reduced